1. How to find the numbers, if there are any, at which f has a realtive minimum. What is the realive minima?
The graph that I have looks like W, it has point at -1 on the x axis, 1 on the y axis, and 1 on the x axis.
Choices that I have:
a. f has a relative minimum at x=0; the relative minimum is 1.
b.f has a relative minimum at x=-1 and 1; the relative minimum is 0.
c.f has a relative minimum at x=-1; the relative minimum is 0.
d.f has no relative minimum.
Determine whether the given ntion is even, odd, or neither (I don't understand how you these problems)
2. f(x)=x^3 -x^2
3. f(x)=-4^5+x^3
To find the numbers, if any, at which a function has a relative minimum, we need to use the first and second derivative tests.
1. Relative minimum of f(x):
- First, we identify the critical points of the function by finding where its derivative equals zero or does not exist.
- Then, we analyze the sign changes of the derivative around those critical points to determine if they correspond to relative minimum points.
Let's start with the given function f(x)=x^3 -x^2.
- First derivative: f'(x) = 3x^2 - 2x.
- Setting f'(x) equal to zero and solving for x, we get: 3x^2 - 2x = 0.
- Factoring out x, we have: x(3x - 2) = 0.
- Therefore, x=0 or x=2/3 are potential critical points.
Now, let's analyze the sign changes of f'(x) around these points:
- For x < 0, we can test a value, e.g., x = -1, in f'(x): f'(-1) = 3(-1)^2 - 2(-1) = 3 +2 = 5 > 0.
- For 0 < x < 2/3, we can test x = 1/2: f'(1/2) = 3(1/2)^2 - 2(1/2) = 3/4 - 1 = -1/4 < 0.
- For x > 2/3, we can try x = 1: f'(1) = 3(1)^2 - 2(1) = 3 - 2 = 1 > 0.
Based on the sign changes, we can conclude:
- At x=0, there is a relative minimum because the derivative changes from positive to negative at that point.
- The relative minimum value is 0, which corresponds to f(0) = (0)^3 - (0)^2 = 0.
Comparing the analysis with the given choices:
- Option (a) states that f has a relative minimum at x = 0 with a value of 1, but our analysis indicates it is incorrect.
- Option (b) states that f has relative minima at x = -1 and 1 with a value of 0, which is also incorrect.
- Option (c) states that f has a relative minimum at x = -1 with a value of 0, which is inconsistent with our findings.
- Option (d) states that f has no relative minimum, which aligns with our conclusion based on the analysis. Therefore, the correct answer is (d).
Moving on to the second question regarding whether a given function is even, odd, or neither, we apply the principles of symmetry:
2. f(x) = x^3 - x^2.
- A function is even if f(-x) = f(x) for all x in the domain. Let's see if this holds for f(x):
f(-x) = (-x)^3 - (-x)^2 = -x^3 - x^2.
Since f(x) = x^3 - x^2 and f(-x) = -x^3 - x^2, we can see that f(-x) ≠ f(x).
- A function is odd if f(-x) = -f(x) for all x in the domain. Let's check if this holds for f(x):
-f(x) = -(x^3 - x^2) = -x^3 + x^2.
As f(-x) = -x^3 - x^2 ≠ -x^3 + x^2 = -f(x), we see that f(-x) ≠ -f(x) as well.
Therefore, the given function f(x) = x^3 - x^2 is neither even nor odd.
To find the numbers at which f has a relative minimum, we need to determine the critical points of the function.
1. First, let's analyze the graph provided. You mentioned that the graph looks like a "W," with a point at (-1, 1) and (1, 0) on the x-axis.
2. The points (-1, 1) and (1, 0) correspond to specific coordinates on the graph, indicating that these points might be critical points.
3. To confirm if these points are relative minima, we have to analyze the behavior of the function around these points.
4. To do so, we can take the derivative of the function, f'(x), and set it equal to 0 to find the critical points.
Now let's solve for the critical points for both functions:
2. For f(x) = x^3 - x^2:
- Calculating the derivative of f(x) gives us f'(x) = 3x^2 - 2x.
- Setting f'(x) = 0, we have 3x^2 - 2x = 0.
- Factoring the equation, we get x(3x - 2) = 0.
- Setting each factor equal to zero, we find x = 0 and x = 2/3.
Therefore, the critical points of f(x) = x^3 - x^2 are x = 0 and x = 2/3.
3. For f(x) = -4x^5 + x^3:
- Taking the derivative of f(x), we get f'(x) = -20x^4 + 3x^2.
- Setting f'(x) = 0, we have -20x^4 + 3x^2 = 0.
- Factoring the equation, we obtain x^2(-20x^2 + 3) = 0.
- Setting each factor equal to zero, we find x = 0 and x = ±√(3/20).
Therefore, the critical points of f(x) = -4x^5 + x^3 are x = 0 and x = ±√(3/20).
Now that we have found the critical points, we can determine if they correspond to relative minima. The question mentions that the relative minima occur when f(x) = 0 or when f(x) = 1. Let's evaluate the function at the critical points to determine the relative minima:
1. For f(x) = x^3 - x^2 (Critical points: x = 0 and x = 2/3):
- Evaluating f(0) gives us f(0) = (0)^3 - (0)^2 = 0.
- Evaluating f(2/3) gives us f(2/3) = (2/3)^3 - (2/3)^2 ≈ 0.09.
From this analysis, we find that f(x) does not equal 1 at any of the critical points. Therefore, the relative minimum is not 1.
2. For f(x) = -4x^5 + x^3 (Critical points: x = 0 and x = ±√(3/20)):
- Evaluating f(0) gives us f(0) = -4(0)^5 + (0)^3 = 0.
- Evaluating f(√(3/20)) gives us f(√(3/20)) = -4(√(3/20))^5 + (√(3/20))^3 ≈ -0.07.
- Evaluating f(-√(3/20)) gives us f(-√(3/20)) = -4(-√(3/20))^5 + (-√(3/20))^3 ≈ -0.07.
Again, we don't find any critical points where f(x) equals 1. Thus, the relative minimum is not 1.
To summarize:
- For f(x) = x^3 - x^2, the function has critical points at x = 0 and x = 2/3. However, the relative minimum is not 1.
- For f(x) = -4x^5 + x^3, the function has critical points at x = 0 and x = ±√(3/20). However, the relative minimum is not 1.
Therefore, the answer is d. f has no relative minimum at any particular x-value.